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5.73 Lecture #11 11 - 1 Eigenvalues, Eigenvectors, and Discrete Variable Representation (DVR) should have read CDTL pages 94-144 Last time: bra (a * 1 * … aN ) ψ in {φ} basis set φ a1 0 a M 2 ψ i = 1 = M M M 0 ψ a N φ a1 M aN φ ket N × N matrix (complex ) # 1 1 1 1 = 0 0 O = ∑k a j= φj ψ i φ φ k k at end of lecture φ i AB φ j = ∑ φ i A φ k φ k B φ j 1 424 3 k 1 = ∑ A ik Bkj = (AB)ij k updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 2 What is the connection between the Schrödinger and Heisenberg representations? ψ i ( x) = x ψ i x 0 = δ( x, x 0 ) eigenfunction of x with eigenvalue x0 Using this formulation for ψi(x), you can go freely (and rigorously) between the Schrödinger and Heisenberg approaches. 1= ∑ k Today: k k = ∫ x x dx eigenvalues of a matrix – what are they? how do we get them? (secular equation). Why do we need them? eigenvectors – how do we get them? Arbitrary V(x) in Harmonic Oscillator Basis Set (DVR) updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 3 Schr. Eq. is an eigenvalue equation Âψ = aψ 0 a1 a2 A= O aN ψ 0 in matrix language A ψi = ai ψi 1 0 ψ1 = 0 M ψ satisfies A ψ 1 = a 1 ψ 1 but that is the eigen-basis representation – a special representation! What about an arbitrary representation? Call it the φ representation. N A c i i φ = a ci i φ * * * * * * i =1 i A as transformation on each φ i ∑ ∑ N unknown coefficients {c i } Eigenvalue equation i = 1 to N How to determine {c i } and a ? Secular Eqn. derive it. first, left multiply by φ j ∑A c φ ji i =a i ∑c N 0= i j i =a i ∑ c [A i φ ji ∑c δ i ij i − aδ ij i =1 ] one equation N unknowns next, multiply original equation by φ k N 0= ∑ c [A i φ ki i =1 etc. for all φ − aδ ik ] another equation . N linear homogeneous equations in N unknowns – Condition that a nontrivial (i.e. not all 0’s) solution exists is that determinant of coefficients = 0. updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 4 A11 − a A12 … A 22 − a O A 21 0= A1N A NN − a Nth order equation – as many as N different values of a satisfy this equation (if fewer than N, some values of a are “degenerate”). Does everyone know how to expand a determinant? {ai} are the eigenvalues of A (same as what we would have obtained by solving differential operator eigenvalue equation) If we know the eigenvalues, then we can find the N { ψ i ψi = ∑ cj j j φ } such that expand the eigenbasis in (computationally convenient) basis ψ i Aψ ψ j = a jδ ij 0 a1 Aψ = 0 O aN A ψ ψ1 1 0 = a1 0 M But we generally start with Aφ in nondiagonal form computer 1. 2. 3. transform to diagonal form by T ≤ Aφ T = A ψ the diagonal elements are eigenvalues the diagonalizing transformation is composed of eigenvectors, column by column of T ≤ . updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 5 Hermitian Matrices A = A† A†ij = A*ji (can use this property to show that all expectation values of A are real) These matrices can be “diagonalized” (i.e. the set of all eigenvalues can be found) by a unitary transformation. T −1 = T ≤ unitary matrix 0 a1 a 2 ≤ φ ≡ Aψ diagonalization T A T = O diagonal not diagonal 0 a N ψ ≤ φ ≤ T A TT φ j diagonal Aψ Eigenvector expressed in φ basis set updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 6 0 0 T1≤i M M ≤ T T ≤ φ i = T ≤ 1 = 2 i = ψ i = 1 ← i - th position M M M ≤ i-th TNi φ position 0 φ 0 ψ eigenvector i-th column of T† suppose we apply Aψ ψ i 0 M = Aψ 1 M i-th 0 ψ 0 M = ai M 0 ψ 0 M = ai 1 M 0 ψ RECAPITULATE: Start with arbitrary basis set φ Construct Aφ : Not Diagonal, but basis set was computationally convenient. a1 0 0 = Aψ O Find T (computer) that causes T A T = aN 0 ≤ φ Eigenstates (eigenkets) are columns of T ≤ in φ basis set. Columns of T are the linear combination of eigenvectors that correspond to each basis state. Useful for “bright state” calculations. updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 7 Can now solve many difficult appearing problems! Start with a matrix representation of any operator that is expressable as a function of a matrix. e.g. e − iH( t −t 0 )/h propagator f ( x) potential curve , prescription example f ( x ) = Tf (T† xT)T† 123 x1 † T xT = 0 diagonalize x – so f( ) is applied to each diagonal element x2 0 O xN f (x1 ) 0 f (x 2 ) † f (T xT) = O 0 f (x N ) Then perform inverse transformation T f(T† x T)T† – undiagonalizes matrix, to give matrix representation of desired function of a matrix. Show that this actually is valid for simple example f(x) = x N f(x) = T T†xT T†xT L T†xT T† [( [ 1 † N )( ] 2 † ) ( = T T x T T = xN N )] apply prescription get expected result general proof for arbitrary f(x) → expand in power series. Use previous result for each integer power. updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 8 John Light: Discrete Variable Representation (DVR) General V(x) evaluated in Harmonic Oscillator Basis Set. we did not do H-O yet, but the general formula for all of the nonzero matrix elements of x is: 1/2 h n x n +1 = 2ωµ (infinite dimension matrix) x 2 = h 2ωµ 1/2 ( n + 1)1/2 ω = (k µ) 1/2 h x= 2ωµ 1 0 2 0 2 0 3 0 6 0 5 0 0 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 1 0 1 0 2 M M 0 6 0 7 0 0 0 0 0 2 L L 3 M 0 3 M M 0 4 L 0 0 L 0 0 0 = 0 0 4 0 0 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 9 0 0 0 0 11 0 0 0 13 0 0 0 15 O 0 0 O O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [CARTOON] etc. matrix multiplication to get matrix for f(x) diagonalize e.g., 1000 × 1000 (truncated) x matrix that was expressed in harmonic oscillator basis set. updated 10/1/02 11:00 AM 5.73 Lecture #11 x1 0 T†xT = 0 0 11 - 9 0 0 0 x2 0 0 0 O 0 0 0 x 1000 x diagonalized-x basis {xi} are eigenvalues. They have no obvious physical significance. 0 0 0 V( x 1 ) 0 0 V( x 2 ) 0 V( x ) x = 0 O 0 0 0 0 0 V( x 1000 ) full † V( x ) H − O = TV( x ) x T = complicated matrix H= next transform back from x-basis to H-O basis set x 1000 T was determined when x was diagonalized × 1000 H−O p2 + V(x) 2µ need matrix for p2, get it from p (the general formula for all non-zero matrix elements of p) updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 10 1/2 h(ωµ) 1/2 n p n + 1 = −i n + 1 ( ) 2 0 1 0 0 2 1/2 − 1 h(ωµ) p = −i − 2 0 0 2 0 O 0 0 O 0 −1 0 2 0 0 −3 0 O h(ωµ) p2 = − 2 0 −5 0 2 0 O O O 0 O O O if H= p2 1 2 + kx 2µ 2 2 hω 0 H= 4 0 0 0 0 0 0 0 O O O O 0 0 O O 0 same structure as x 1 k = 1 ω 2 µ 2 2 0 0 0 1 / 2 0 0 0 0 0 6 0 0 = hω 0 3 / 2 0 10 0 0 0 5 / 2 0 0 0 14 0 0 O 0 but for arbitrary V(x), express H in HO basis set, 2 pHO H HO = + V( x ) HO 2µ TV(x)xT† E1 0 0 eigenvalues obtained by S H HOS = 0 O 0 0 0 EN † columns of S† are eigenvectors in HO basis set! updated 10/1/02 11:00 AM 5.73 Lecture #11 11 - 11 1. Express matrix of x in H-O basis (automatic; easy to program a computer to do this), get xHO. 2. Diagonalize xHO. Get xx and T. 3. Trivial to write V(x)x as V(xi) = V(x)x in x basis 4. Transform V(x)x back to V(x)HO 5. Diagonalize HHO. Solve many V(x) problems in this basis set. 1000 × 1000 T matrix diagonalizes x ⇒ 1000 xi’s Save the T and the {xi} for future use on all V(x) problems. To verify convergence, repeat for new x matrix of dimension 1100 x 1100. There will be no resemblance between {x i}1000 and {x i}1100 . If the lowest eigenvalues of H (i.e. the ones you care about) do not change (by measurement accuracy), converged! Next: Matrix solution of HO (no wave functions at all) Start from Commutation Rule! Then Perturbation Theory. updated 10/1/02 11:00 AM